Problem Analysis: Triangle Area Calculation

This problem requires calculating the area of a triangle given the coordinates of one vertex (A), the constraints for two other vertices (B and C) lying on a specific line, and the distance between B and C.

Given Information

• Vertex A: $(1, 6, 3)$
• Vertex B: $(4, 9, \alpha)$
• Vertices B and C lie on the line: $\frac{x}{1} = \frac{1 - y}{-2} = \frac{z - 2}{3}$
• Distance BC = 10 units
• Goal: Find the area of $\Delta ABC$.

Line Equation Analysis

The given line equation can be rewritten in standard symmetric form:

$ \frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3} $

From this form, we identify:

• A point on the line, $P_0 = (0, 1, 2)$.
• The direction vector of the line, $\vec{d} = \langle 1, 2, 3 \rangle$.

Finding Coordinate B

Since vertex B $(4, 9, \alpha)$ lies on the line, its coordinates must satisfy the line equation:

$ \frac{4}{1} = \frac{9 - 1}{2} = \frac{\alpha - 2}{3} $

Evaluating the first two parts:

$ 4 = \frac{8}{2} = 4 $

Now, equating the first part with the third:

$ 4 = \frac{\alpha - 2}{3} $

Solving for $\alpha$:

$ 12 = \alpha - 2 $

$ \alpha = 14 $

Therefore, the coordinates of vertex B are $(4, 9, 14)$.

Calculating Triangle Area

The area of $\Delta ABC$ can be calculated using the formula: Area $= \frac{1}{2} \times \text{base} \times \text{height}$.

• Base: The length of the base BC is given as 10 units.
• Height: The height is the perpendicular distance from vertex A to the line containing the base BC.

To find the height ($h$), we use the formula:

$ h = \frac{|\vec{AP_0} \times \vec{d}|}{|\vec{d}|} $

Where $P_0$ is a point on the line and $\vec{d}$ is the direction vector of the line.

Step 1: Find vector $\vec{AP_0}$

Using $A = (1, 6, 3)$ and $P_0 = (0, 1, 2)$:

$ \vec{AP_0} = P_0 - A = \langle 0 - 1, 1 - 6, 2 - 3 \rangle = \langle -1, -5, -1 \rangle $

Step 2: Calculate the cross product $\vec{AP_0} \times \vec{d}$

$ \vec{AP_0} \times \vec{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & -5 & -1 \\ 1 & 2 & 3 \end{vmatrix} $

$ = \mathbf{i}((-5)(3) - (-1)(2)) - \mathbf{j}((-1)(3) - (-1)(1)) + \mathbf{k}((-1)(2) - (-5)(1)) $

$ = \mathbf{i}(-15 + 2) - \mathbf{j}(-3 + 1) + \mathbf{k}(-2 + 5) $

$ = -13\mathbf{i} + 2\mathbf{j} + 3\mathbf{k} = \langle -13, 2, 3 \rangle $

Step 3: Calculate the magnitudes

Magnitude of the cross product:

$ |\vec{AP_0} \times \vec{d}| = \sqrt{(-13)^2 + 2^2 + 3^2} = \sqrt{169 + 4 + 9} = \sqrt{182} $

Magnitude of the direction vector:

$ |\vec{d}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14} $

Step 4: Calculate the height

$ h = \frac{\sqrt{182}}{\sqrt{14}} = \sqrt{\frac{182}{14}} = \sqrt{13} $

Step 5: Calculate the Area

$ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \times \sqrt{13} $

$ \text{Area} = 5\sqrt{13} $

Based on these calculations, the area is $5\sqrt{13}$ square units, which corresponds to Option 3.

However, the provided correct answer is Option B: $15\sqrt{13}$.